Cryptarithmetic Practice Set for eLitmus and Infosys: 2026 Edition

Cryptarithmetic replaces each digit in an arithmetic equation with a letter, and the task is to find which digit each letter represents.

Two aptitude tests that include this puzzle type are the Infosys online test and the eLitmus pH Test, which features cryptarithmetic in its problem-solving module. The six worked problems below are ordered by difficulty: two simple 2-digit additions, two intermediate puzzles that reveal structural patterns, the canonical SEND + MORE = MONEY in full, and one multiplication variant that tests a different approach.

Four Ground Rules for Any Cryptarithmetic Puzzle

Before starting, confirm the puzzle satisfies these four conditions. A problem that breaks any of them is either unsolvable or has more than one answer.

• Unique digits. Each letter maps to exactly one digit (0-9). No two letters share a digit.
• No leading zeros. The leftmost letter of any multi-digit number cannot equal 0. In SEND + MORE = MONEY, neither S nor M can be 0.
• Exactly one solution. A well-formed puzzle has one valid assignment. Two solutions indicate a missing constraint or an exam question that asks for a common property across both.
• Arithmetic holds. Once digits are substituted, the equation is numerically correct. This verification step catches errors that crept in during analysis.

The Core Technique: Column-by-Column Carry Analysis

Three principles drive the approach:

• Binary carries. Carries between columns are 0 or 1. Combined with digit uniqueness, this converts what appears to be a 9! (nine-factorial = 362,880 possible assignments for 9 distinct letters) search into two to four test cases.
• Start at the leading column. The leftmost column of the result often pins a carry value directly. A carry that generates an extra digit is always 1.
• Work right to left. Track which carry values are forced at each column before testing digit assignments.

Simple Problems: The Carry Method in Action

Problem 1 — TO + GO = OUT

  T O
+ G O
-----
O U T

Every letter is a unique digit. T and G are leading digits, so neither equals 0. Solve for T, O, G, U.

• Step 1 — Hundreds digit of the result. OUT is a 3-digit result from two 2-digit numbers. A 3-digit result requires a carry into the hundreds position. That carry equals 1. O is the hundreds digit of OUT, so O = 1.
• Step 2 — Ones column. O + O = T + 10 times c1, where c1 is the carry to the tens column. Substituting O = 1 gives 2 = T + 10 times c1. Since T is a single digit and 2 is less than 10, c1 = 0 and T = 2.
• Step 3 — Tens column. T + G + c1 = U + 10 times 1 (the hundreds carry). Substituting T = 2 and c1 = 0 gives G minus U = 8. The only single-digit pair with this difference that avoids O = 1 and T = 2 is G = 8, U = 0. (G = 9, U = 1 conflicts with O = 1.)
• Solution: T = 2, O = 1, G = 8, U = 0.
• Verification: 21 + 81 = 102, matching OUT with O = 1, U = 0, T = 2. Unique solution.

Problem 2 — AB + BA = CDC

CDC means C in the hundreds place, D in the tens place, C in the units place (C repeats).

• Step 1 — Algebraic setup. AB = 10A + B, BA = 10B + A. Their sum is 11(A + B). CDC = 101C + 10D. So 11(A + B) = 101C + 10D.
• Step 2 — Pin C. The sum of two 2-digit numbers is at most 189. So CDC is at most 189, meaning C = 1 (C is a leading digit and cannot be 0).
• Step 3 — Pin D. Substituting C = 1: 11(A + B) = 101 + 10D. For the left side to be divisible by 11, check (101 + 10D) mod 11. Since 101 equals 9 times 11 plus 2, we have 101 mod 11 = 2, and 10 mod 11 = 10. So (2 + 10D) mod 11 = 0 gives D = 2.
• Step 4 — Pin A + B. 11(A + B) = 101 + 20 = 121, so A + B = 11.
• Solution: C = 1, D = 2 are fixed. A and B can be any distinct pair summing to 11, where neither equals 1 or 2: (3, 8), (4, 7), (5, 6), and their reverses.
• Verification: A = 3, B = 8: 38 + 83 = 121. A = 4, B = 7: 47 + 74 = 121. Both confirm CDC = 121. C and D are uniquely determined; A and B are not.

Intermediate Problems: Parity and Structural Patterns

Problem 3 — ODD + ODD = EVEN

EVEN has E in the thousands and tens positions (it repeats), V in the hundreds, N in the units.

  O D D
+ O D D
-------
E V E N

• Step 1 — Pin E. ODD + ODD = 2 times ODD. The maximum is 2 times 999 = 1998. The minimum 4-digit result from this form requires ODD at least 500. So E = 1 (leading digit), and the carry c3 out of the hundreds column equals 1.
• Step 2 — Tens column. 2D + c1 = E + 10 times c2 = 1 + 10 times c2.
  • c2 = 0: 2D + c1 = 1. Only D = 0, c1 = 1. Units: 2 times 0 = N + 10 gives N = -10. Impossible.
  • c2 = 1: 2D + c1 = 11. Only D = 5, c1 = 1 (2 times 5 + 1 = 11).
• Step 3 — Units column. With c1 = 1: 2 times 5 = N + 10, so N = 0.
• Step 4 — Hundreds column. With c2 = 1 and c3 = 1: 2O + 1 = V + 10, so V = 2O minus 9. For V at least 0, O is at least 5. Excluding O = 5 (equals D), O = 1 (equals E), and O = 0 (equals N):
  • O = 6: V = 3. Digits {E=1, D=5, N=0, O=6, V=3} are all distinct. 655 + 655 = 1310. ✓
  • O = 7: V = 5 = D. Conflict.
  • O = 8: V = 7. Digits {E=1, D=5, N=0, O=8, V=7} are all distinct. 855 + 855 = 1710. ✓
  • O = 9: V = 9 = O. Conflict.
• Two valid solutions: 655 + 655 = 1310 and 855 + 855 = 1710. D = 5, E = 1, and N = 0 are identical in both. An exam question asking for D, E, or N has a unique answer regardless of which solution holds.

Problem 4 — ABAB + BABA = CCCC

• Step 1 — Factor ABAB. ABAB = 1000A + 100B + 10A + B = 1010A + 101B = 101 times (10A + B).
• Step 2 — Factor BABA. By symmetry, BABA = 101 times (10B + A).
• Step 3 — Sum. ABAB + BABA = 101 times (10A + B + 10B + A) = 101 times 11 times (A + B) = 1111 times (A + B).
• Step 4 — CCCC. CCCC = 1111 times C. So A + B = C.
• Constraints: A at least 1 (leading digit of ABAB), B at least 1 (leading digit of BABA), A not equal to B, and C = A + B must be a single digit, so A + B at most 9.
• Examples: A = 1, B = 2, C = 3: 1212 + 2121 = 3333. A = 4, B = 5, C = 9: 4545 + 5454 = 9999. Both verified. ✓
• Key insight: Algebraic factoring collapses what looks like a multi-column problem into one constraint. When you see a repeating-digit pattern in the result, factoring often beats column-by-column analysis.

The Flagship Puzzle: SEND + MORE = MONEY

This is the most cited cryptarithmetic example in placement preparation. The full derivation uses carry analysis across 8 steps.

  S E N D
+ M O R E
---------
M O N E Y

Distinct letters: {S, E, N, D, M, O, R, Y}. Eight letters, eight unique digits from 0 to 9.

• Step 1. MONEY is a 5-digit result from two 4-digit addends. The only carry that generates a fifth digit is 1. So M = 1.
• Step 2 — Thousands column. S + M + c3 = O + 10 times M. With M = 1: S + 1 + c3 = O + 10, so S + c3 = O + 9.
  • c3 = 0: S minus O = 9. Only pair with difference 9: S = 9, O = 0.
  • c3 = 1: S minus O = 8. Either S = 9, O = 1 (conflicts with M = 1), or S = 8, O = 0.
• Step 3 — Eliminate c3 = 1. Test the S = 8, O = 0 branch with the hundreds column: E + 0 + c2 = N + 10 times 1. Maximum of E + c2 is 9 + 1 = 10, giving N = 0 = O. Conflict. Discard the c3 = 1 branch.
• Step 4. Confirmed: S = 9, O = 0, c3 = 0.
• Step 5 — Hundreds column. E + 0 + c2 = N (since c3 = 0). So N = E + c2.
• Step 6 — Tens column. N + R + c1 = E + 10 times c2. Substituting N = E + c2 gives R + c1 = 9 times c2.
  • c2 = 0: R + c1 = 0, so R = 0 = O. Conflict.
  • c2 = 1: R + c1 = 9. S = 9 is already taken, so R is not 9. Therefore R = 8, c1 = 1. And N = E + 1.
• Step 7 — Units column. D + E = Y + 10 times c1 = Y + 10. So Y = D + E minus 10.
• Step 8 — Assign remaining digits. Digits used: M = 1, S = 9, O = 0, R = 8. Remaining pool for E, N = E + 1, D, Y: {2, 3, 4, 5, 6, 7}. Need D + E at least 10 for Y to be non-negative.
  • E = 5, N = 6: D from pool excluding 5 and 6 is {2, 3, 4, 7}. D = 7 gives Y = 2. Both 7 and 2 are in the pool. Digits {M=1, O=0, R=8, S=9, E=5, N=6, D=7, Y=2} are all distinct. Valid.
  • E = 2, N = 3: D must be at least 8; no digit at least 8 remains in the pool. Ruled out.
  • E = 3, N = 4: D = 7 gives Y = 0 = O. Conflict.
  • E = 4, N = 5: D = 6 gives Y = 0 = O. D = 7 gives Y = 1 = M. Both conflict.
  • E = 6, N = 7: D = 4 gives Y = 0 = O. D = 5 gives Y = 1 = M. Both conflict.
• Solution: M = 1, O = 0, R = 8, S = 9, E = 5, N = 6, D = 7, Y = 2.
• Verification: SEND = 9567, MORE = 1085. 9567 + 1085 = 10652 = MONEY. Unique solution. ✓

The Multiplication Variant: ABCD times 4 = DCBA

A 4-digit number multiplied by 4 gives the same four digits in reverse order.

  A B C D
×       4
---------
  D C B A

• Step 1 — Bound A. The result DCBA is 4 digits, so ABCD times 4 is at most 9999, meaning ABCD is at most 2499. Therefore A is 1 or 2.
• Step 2 — Ones column. The ones digit of ABCD times 4 equals A (the ones digit of DCBA). So (4 times D) mod 10 = A.
  • A = 1: 4D mod 10 = 1. Impossible since 4D is always even.
  • A = 2: 4D mod 10 = 2, giving D = 3 or D = 8.
• Step 3 — Test D = 3. Equation: 4 times (2BC3) = 3CB2. Expanding: 8000 + 400B + 40C + 12 = 3002 + 100C + 10B. This gives 390B minus 60C = -5010. No valid single-digit B and C satisfy this.
• Step 4 — Test D = 8. Equation: 4 times (2BC8) = 8CB2. Expanding: 8000 + 400B + 40C + 32 = 8002 + 100C + 10B. This gives 390B minus 60C = -30, or 13B minus 2C = -1, so 2C = 13B + 1.
  • B = 1: C = 7. Digits A = 2, B = 1, C = 7, D = 8 are all distinct. Valid.
  • B = 0: C = 0.5 (not an integer). B = 3: C = 20 (not a single digit). No other valid B.
• Solution: A = 2, B = 1, C = 7, D = 8. ABCD = 2178.
• Verification: 2178 times 4 = 8712. Column by column: 8 times 4 = 32 (write 2, carry 3); 7 times 4 = 28 + 3 = 31 (write 1, carry 3); 1 times 4 = 4 + 3 = 7; 2 times 4 = 8. Result: 8712 = DCBA. Unique solution. ✓

Where These Puzzles Appear in Infosys and eLitmus Tests

Infosys screens freshers through three tracks with different aptitude demands.

Track	Starting CTC	Aptitude difficulty
System Engineer (SE)	₹3.6 LPA	Standard aptitude: series, analogies, basic deduction
Specialist Programmer (SP)	₹6.5 LPA	Stronger logical deduction; InfyTQ certification preferred
Power Programmer (PP)	₹9.5 LPA	HackWithInfy performance; advanced reasoning required

Cryptarithmetic-type questions appear most often at the SP and PP difficulty levels. The SE track covers standard aptitude without puzzle-heavy logical variants.

According to Q4 FY26 commentary from Infosys CEO Salil Parekh, Infosys offers different starting compensation for candidates with AI-attuned skills and is building a pool of forward-deployed engineers to work directly on AI solution projects with clients. Infosys onboarded 20,000 freshers in FY26 and plans similar numbers in FY27. The harder aptitude tracks are where that pool gets identified.

On the eLitmus side, the eLitmus pH Test complete guide covers the full syllabus, exam pattern, and preparation approach. Cryptarithmetic appears in the Reasoning section under tighter time pressure than the Infosys version. For broader Infosys aptitude practice, the Infosys placement papers collection and the number series guide cover the full aptitude section range alongside cryptarithmetic.

The carry-analysis technique and the two-solution outcome in ODD + ODD = EVEN both demonstrate the same underlying skill: narrowing a search space by eliminating invalid assignments before enumerating candidates. That approach transfers to AI work, particularly in prompt design and model evaluation. TinkerLLM at ₹299 gives engineering students hands-on time with production LLM APIs, starting with exactly that constraint-first thinking. The AI engineering roadmap breaks down the curriculum without the enrolment commitment.